Question

Consider the titration of a 22.0 −mL sample of 0.105 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.

Part E

the pH at the equivalence point Express your answer using two decimal places.

Part F

the pH after adding 6.00 mL of base beyond the equivalence point Express your answer using two decimal places.

Answer 1

E)

find the volume of NaOH used to reach equivalence point

M(HC2H3O2)*V(HC2H3O2) =M(NaOH)*V(NaOH)

0.105 M *22.0 mL = 0.13M *V(NaOH)

V(NaOH) = 17.7692 mL

Given:

M(HC2H3O2) = 0.105 M

V(HC2H3O2) = 22 mL

M(NaOH) = 0.13 M

V(NaOH) = 17.7692 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.105 M * 22 mL = 2.31 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.13 M * 17.7692 mL = 2.31 mmol

We have:

mol(HC2H3O2) = 2.31 mmol

mol(NaOH) = 2.31 mmol

2.31 mmol of both will react to form C2H3O2- and H2O

C2H3O2- here is strong base

C2H3O2- formed = 2.31 mmol

Volume of Solution = 22 + 17.7692 = 39.7692 mL

Kb of C2H3O2- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofC2H3O2-,c = 2.31 mmol/39.7692 mL = 0.0581M

C2H3O2- dissociates as

C2H3O2- + H2O -----> HC2H3O2 + OH-

0.0581 0 0

0.0581-x x x

Kb = [HC2H3O2][OH-]/[C2H3O2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-10)*5.809*10^-2) = 5.681*10^-6

since c is much greater than x, our assumption is correct

so, x = 5.681*10^-6 M

[OH-] = x = 5.681*10^-6 M

use:

pOH = -log [OH-]

= -log (5.681*10^-6)

= 5.2456

use:

PH = 14 - pOH

= 14 - 5.2456

= 8.7544

Answer: 8.75

F)

Given:

M(HC2H3O2) = 0.105 M

V(HC2H3O2) = 22 mL

M(NaOH) = 0.13 M

V(NaOH) = 6 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.105 M * 22 mL = 2.31 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.13 M * 6 mL = 0.78 mmol

We have:

mol(HC2H3O2) = 2.31 mmol

mol(NaOH) = 0.78 mmol

0.78 mmol of both will react

excess HC2H3O2 remaining = 1.53 mmol

Volume of Solution = 22 + 6 = 28 mL

[HC2H3O2] = 1.53 mmol/28 mL = 0.0546M

[C2H3O2-] = 0.78/28 = 0.0279M

They form acidic buffer

acid is HC2H3O2

conjugate base is C2H3O2-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {2.786*10^-2/5.464*10^-2}

= 4.452

Answer: 4.45